Lol
Are you familiar with what Mathgician said? Its a well known Trigonometric Identity, [itex]\tan^2 x = \sec^2 x -1[/itex]
Knowing that,
[tex]\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \int \sec^2 x dx - \int 1 dx = (\int \sec^2 x dx ) - x[/tex]
For the integral of (sec x) squared, if you don't already know it, try differentiating tan x, what is that?